#1

Hey guys,

I'm retarded. Seriously, this stuff has gone right over my head.

The problem is :

A body is thrown upwards with an initial speed 'u' from a point 'P' which is 40m above the ground. After 6 seconds it is at a point 'Q' and its velocity 8m/s downwards. Find the value of 'u' and the height of 'Q' above thr ground.

The answers are meant to be 50.8m/s and 168.4m, can somebody explain this to me please ?

Thanks a lot UG

I'm retarded. Seriously, this stuff has gone right over my head.

The problem is :

A body is thrown upwards with an initial speed 'u' from a point 'P' which is 40m above the ground. After 6 seconds it is at a point 'Q' and its velocity 8m/s downwards. Find the value of 'u' and the height of 'Q' above thr ground.

The answers are meant to be 50.8m/s and 168.4m, can somebody explain this to me please ?

Thanks a lot UG

#2

The answer is 7

#3

The answer is 7

No, but then you get the divide by zero error and oh shi-

#4

uhhhhhhhhhhhhhhhhhhhh

im in biology

im in biology

#5

Physics was such a long time ago...

Does it help that the acceleration on Earth is 9.8 m/s2? I had to do problems like that all the time and it helped to know the rate at which an object accelerates in free fall.

Does it help that the acceleration on Earth is 9.8 m/s2? I had to do problems like that all the time and it helped to know the rate at which an object accelerates in free fall.

#6

If you hang tight, I (or someone else) will figure this out.

#7

velocity^2=2(change in position times acceleration)+initial velocity.

since the body is going straight up in the air, the acceleration is -9.8m/s^2 due to the earth's gravity.

voila.

since the body is going straight up in the air, the acceleration is -9.8m/s^2 due to the earth's gravity.

voila.

#8

By the way, are you neglecting friction due to air? Hope so.

#9

By the way, are you neglecting friction due to air? Hope so.

We are neglecting it, yeah.

Ugh, obviously I'm not mathematically minded at all ! Can't wrap my head around these things even slightly.

Thanks a lot for the help though guys

#10

uhhhhhhhhhhhhhhhhhhhh

im in biology

HIGH FIVE!

#11

velocity^2=2(change in position times acceleration)+initial velocity.

since the body is going straight up in the air, the acceleration is -9.8m/s^2 due to the earth's gravity.

voila.

That's my issue.... I can't remember the kinematics formulas. I give then.

#12

The formulae are :

v=u+at

s=ut+1/2at

v^2=u^2+2as

(Taking:

u=initial velocity

v=final velocity

a=acceleration

s=displacement

t=time)

If that makes sense, they're probably taught differently everywhere though!

v=u+at

s=ut+1/2at

v^2=u^2+2as

(Taking:

u=initial velocity

v=final velocity

a=acceleration

s=displacement

t=time)

If that makes sense, they're probably taught differently everywhere though!

#13

velocity^2=2(change in position times acceleration)+initial velocity.

since the body is going straight up in the air, the acceleration is -9.8m/s^2 due to the earth's gravity.

voila.

the initial velocity should be squared also.

once you plug everything in it should be 64 = v0^2 + 2*9.8*(Q-40) but thats not enough to solve the equation. use the other equation: x1 = x0 +v0*t + 1/2*a*t^2 and plug everthing in for that and you get Q = 40 + v0*6 + 1/2*9.8*36.

use those 2 equations and solve for v0

#14

the initial velocity should be squared also.

once you plug everything in it should be 64 = v0^2 + 2*9.8*(Q-40) but thats not enough to solve the equation. use the other equation:x1 =x0 +v0*t + 1/2*a*t^2 and plug everthing in for that and you get Q = 40 + v0*6 + 1/2*9.8*36.

use those 2 equations and solve for v0

Cool, ok thanks a lot, but where did you get the two x's I bolded ? What do they represent ?

Thanks !

#15

x is another way to say position. (delta)x = change in distance, or total displacement as you called it somewhere. His x corresponds to your s.

#16

Hey guys,

I'm retarded. Seriously, this stuff has gone right over my head.

The problem is :

A body is thrown upwards with an initial speed 'u' from a point 'P' which is 40m above the ground. After 6 seconds it is at a point 'Q' and its velocity 8m/s downwards. Find the value of 'u' and the height of 'Q' above thr ground.

The answers are meant to be 50.8m/s and 168.4m, can somebody explain this to me please ?

Thanks a lot UG

What grade or level of high school/colledge are you in??? ThaT ALL SOUNDS CONFUZING WITH THE LETTERS AND THINGS...BUT I'M IN 8TH AND only doin algebra so i don't know any of this....(sorry for tha caps didn't relize i was doing it...)

#17

What grade or level of high school/colledge are you in??? ThaT ALL SOUNDS CONFUZING WITH THE LETTERS AND THINGS...BUT I'M IN 8TH AND only doin algebra so i don't know any of this....(sorry for tha caps didn't relize i was doing it...)

Then try the BACKSPACE KEY. (That's "delete" on a Mac.) You are allowed to go back and erase what you wrote before you hit the Post button.

#18

Make a list of all the variables, and put in the ones you know.

u = ?

v = -8 (negative, as I'm taking upwards to be the positive direction)

x = ?

t = 6

a = -9.8

Then pick an equation that will find one of the unknowns:

v = u + at

-8 = u - 58.8

u = 50.8 m/s

Then using that result, work out the other one:

v^2 = u^2 + 2ax

64 = 2580.64 -19.6x

19.6x = 2516.64

x = 128.4 m

x is the displacement, so the total distance from the ground, Q = 128.4 + 40 = 168.4m

BISH BASH BOSH!

u = ?

v = -8 (negative, as I'm taking upwards to be the positive direction)

x = ?

t = 6

a = -9.8

Then pick an equation that will find one of the unknowns:

v = u + at

-8 = u - 58.8

u = 50.8 m/s

Then using that result, work out the other one:

v^2 = u^2 + 2ax

64 = 2580.64 -19.6x

19.6x = 2516.64

x = 128.4 m

x is the displacement, so the total distance from the ground, Q = 128.4 + 40 = 168.4m

BISH BASH BOSH!

#19

i know i can do that i was being lazy...sorry

#20

First write down all the values of v, u, a, s and t

v=8

u=?

a=9.8

s=?

t=6

Then look at the formulae to see which of them you can use.

First part:

v=u+at

u=v-at

u=8-(9.8*6)

u= -50.8m/s (negative in this case means upwards)

Second part:

s=ut+0.5at^2

s=-50.8*6 + 9.8 * 36

s= -304.8 + 176.4

s= -128.4m (again, negative is up)

Add that t the starting height, 40m, and you get 168.4m

EDIT: Damn you mynamewontfit >=[

v=8

u=?

a=9.8

s=?

t=6

Then look at the formulae to see which of them you can use.

First part:

v=u+at

u=v-at

u=8-(9.8*6)

u= -50.8m/s (negative in this case means upwards)

Second part:

s=ut+0.5at^2

s=-50.8*6 + 9.8 * 36

s= -304.8 + 176.4

s= -128.4m (again, negative is up)

Add that t the starting height, 40m, and you get 168.4m

EDIT: Damn you mynamewontfit >=[

#21

Oh righty, of course.

So

u=?

t=6

a=-9.8

v=-8

s=0

v=u+at

Sub in and you get 50.8, then take your answer and put it into

s=ut+1/2at^2

and it should work out.

OK, I think I have it. Thanks a lot guys!

Edit : LOL!! 3rd place !! ^^

Thanks loads guys, really helpful !

So

u=?

t=6

a=-9.8

v=-8

s=0

v=u+at

Sub in and you get 50.8, then take your answer and put it into

s=ut+1/2at^2

and it should work out.

OK, I think I have it. Thanks a lot guys!

Edit : LOL!! 3rd place !! ^^

Thanks loads guys, really helpful !

#22

...Close me?